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(Vì x > 0 nên |x| = x; y2 > 0 với mọi y ≠ 0)
(Vì x2 ≥ 0 với mọi x; và vì y < 0 nên |2y| = – 2y)
(Vì x < 0 nên |5x| = – 5x; y > 0 nên |y3| = y3)
(Vì x2y4 = (xy2)2 > 0 với mọi x ≠ 0, y ≠ 0)
a) 1/y
b) - x^2 y
c) -25x^2 / y^2
d) 4x/5y
a) yx.√x2y4=yx.√x2√y4=yx.|x||y2|=yx.xy2=1yyx.x2y4=yx.x2y4=yx.|x||y2|=yx.xy2=1y.
(Do x>0x>0 nên
a) yx.√x2y4=yx.√x2√y4=yx.|x||y2|=yx.xy2=1yyx.x2y4=yx.x2y4=yx.|x||y2|=yx.xy2=1y.
(Do x>0x>0 nên
a) \dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{\sqrt{x^2}}{\sqrt{y^4}}=\dfrac{y}{x}.\dfrac{|x|}{|y^2|}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{1}{y}xy.y4x2=xy.y4x2=xy.∣y2∣∣x∣=xy.y2x=y1.
(Do x>0x>0 nên |x|=x∣x∣=x, y \ne 0y=
a)\(\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}\cdot\dfrac{\left|x\right|}{y^2}=\dfrac{y}{x}\cdot\dfrac{x}{y^2}=\dfrac{1}{y}\)
b)\(2y^2\sqrt{\dfrac{x^4}{4y^2}}=2y^2\dfrac{x^2}{2y}=\dfrac{2y^2\cdot x^2}{2y}=yx^2\)
c)\(5xy\cdot\sqrt{\dfrac{25x^2}{y^6}}=5xy\cdot\dfrac{\sqrt{25x^2}}{\sqrt{y^6}}=5xy\cdot\dfrac{5x}{\left|y^3\right|}=\dfrac{25x^2y}{\left|y^3\right|}\)
a) \(\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{1}{y}\)
b) \(2y^2.\sqrt{\dfrac{x^4}{4y^2}}=2y^2.\dfrac{x^{^2}}{-2y}-x^2y\)
c) \(5xy.\sqrt{\dfrac{25x^2}{y^6}}=5xy.\dfrac{5.\left(-x\right)}{y^3}=\dfrac{-25x^2}{y^2}\)
d) \(0,2x^3y^3.\sqrt{\dfrac{16}{x^4y^8}}=0,2x^3y^3.\dfrac{4}{x^2y^4}=\dfrac{4x}{5y}\)
a) \dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{\sqrt{x^2}}{\sqrt{y^4}}=\dfrac{y}{x}.\dfrac{|x|}{|y^2|}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{1}{y}xy.y4x2=xy.y4x2=xy.∣y2∣∣x∣=xy.y2x=y1.
(Do x>0x>0 nên |x|=x∣x∣=x, y \ne 0y=
Đúng(0)
a) yx.√x2y4=yx.√x2√y4=yx.|x||y2|=yx.xy2=1yyx.x2y4=yx.x2y4=yx.|x||y2|=yx.xy2=1y.
(Do x>0x>0 nên
a) yx.√x2y4=yx.√x2√y4=yx.|x||y2|=yx.xy2=1yyx.x2y4=yx.x2y4=yx.|x||y2|=yx.xy2=1y.
(Do x>0x>0 nên
(Vì x > 0 nên |x| = x; y2 > 0 với mọi y ≠ 0)
(Vì x2 ≥ 0 với mọi x; và vì y < 0 nên |2y| = – 2y)
(Vì x < 0 nên |5x| = – 5x; y > 0 nên |y3| = y3)
(Vì x2y4 = (xy2)2 > 0 với mọi x ≠ 0, y ≠ 0)
a, \(\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\sqrt{\dfrac{x^2}{\left(y^2\right)^2}}=\dfrac{y}{x}.\dfrac{\left|x\right|}{\left|y^2\right|}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{y}{y^2}=\dfrac{1}{y}\)(vì x>0,y≠0)
b, \(2y^2.\sqrt{\dfrac{x^4}{4y^2}}=2y^2.\dfrac{\sqrt{x^4}}{\sqrt{4y^2}}=2y^2.\dfrac{\sqrt{\left(x^2\right)^2}}{\sqrt{\left(2y\right)^2}}=2y^2.\dfrac{\left|x^2\right|}{\left|2y\right|}=2y^2.\dfrac{x^2}{\left(-2y\right)}=2y^2.\dfrac{x^2}{-2y}=\dfrac{y.x^2}{-1}=-x^2y\)
c
\(5xy.\sqrt{\dfrac{25x^2}{y^6}}=5xy.\dfrac{\sqrt{25x^2}}{\sqrt{y^6}}=5xy.\dfrac{\sqrt{\left(5x\right)^2}}{\sqrt{\left(y^3\right)^2}}=5xy.\dfrac{5\left|x\right|}{\left|y^3\right|}=5xy.\dfrac{-5x}{y^3}=\dfrac{-25x^2y}{y^3}=\dfrac{-25x^2}{y^2}\)
d,\(0,2x^3y^3.\sqrt{\dfrac{16}{x^4y^8}}=0,2x^3y^3.\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}=0,2x^3y^3.\dfrac{\sqrt{4^2}}{\sqrt{\left(x^2y^4\right)^2}}=0,2x^3y^3.\dfrac{4}{\left|x^2y^4\right|}=0,2x^3y^3.\dfrac{4}{x^2y^4}=\dfrac{0,8x^3y^3}{x^2y^4}=\dfrac{0,8x}{y}\)
a, \(\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{x}{y^2}\left(vix>0\right)=\dfrac{1}{y}\)
b,
\(a,\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=2y^2.\dfrac{\sqrt{x^4}}{\sqrt[]{4y^2}}=2y^2.\dfrac{\left|x^2\right|}{\left|2y\right|}=2y^2.\dfrac{x^2}{-2y}=\dfrac{2y^2.x^2}{-2y}=-x^2y\)
b,\(2y^2.\sqrt{\dfrac{x^4}{4y^2}}=2y^2.\dfrac{\sqrt{x^4}}{\sqrt{4y^2}}=2y^2.\dfrac{\left|x^2\right|}{\left|-2y\right|}\dfrac{2y^2.x^2}{-2y}=-x^2y\)
c,\(5xy.\sqrt{\dfrac{25x^2}{y^6}}=5xy.\dfrac{\sqrt{25x^2}}{\sqrt{y^6}}=5xy.\dfrac{\left|5x\right|}{\left|y^3\right|}=5xy.\dfrac{\left|-5x\right|}{\left|y^3\right|}=\dfrac{5xy.\left(-5x\right)}{y^3}=\dfrac{-25x^2}{y^2}\)
d,\(0,2x^3.y^3.\sqrt{\dfrac{16}{x^4y^8}}=0,2x^3.y^3.\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}=\dfrac{2}{10}.x^3y^3.\dfrac{4}{\left|x^2y^4\right|}=\dfrac{1}{5}x^3y^3.\dfrac{4}{x^2y^4}\)
a, =\(\dfrac{y}{x}\) . \(\sqrt{x^2}\) /\(\sqrt{\left(y^2\right)^2}\) =\(\dfrac{y}{x}\) . \(\dfrac{\left|x\right|}{\left|y^2\right|}\) = \(\dfrac{y}{x}\) .\(\dfrac{x}{y^2}\) = \(\dfrac{y}{y^2}\)=\(\dfrac{1}{y}\)
b, = 2y2 . \(\sqrt{\left(x^2\right)^2}\) / \(\sqrt{4y^2}\) = 2y2 .\(\dfrac{\left|x^2\right|}{2\left|y\right|}\) = 2y2 .\(\dfrac{x^2}{-2.y}\) = \(\dfrac{y.x^2}{-1}\) = -x2y
c , = 5xy . \(\sqrt{25x^2}\) / \(\sqrt{\left(y^3\right)^2}\) = 5xy . \(\dfrac{5.\left|x\right|}{\left|y^3\right|}\) =5xy . \(\dfrac{-5x}{y^3}\) = \(\dfrac{-25x^2y}{y^3}\) = \(\dfrac{-25x^2}{y^2}\)
d , =0,2x3y3 . \(\sqrt{16}\) / \(\sqrt{x^4y^8}\) = 0,2 x3y3 . \(\dfrac{4}{\sqrt{\left(x^2y^4\right)^2}}\) = 0,2x3y3 .\(\dfrac{4}{\left|x^2y^4\right|}\) = 0,2x3y3 .\(\dfrac{4}{x^2y^4}\)( vì x2y4 ≥0) = \(\dfrac{0,8x^3y^3}{x^2y^4}\) =\(\dfrac{0,8x}{y}\)
a) yx.√x2y4=yx.√x2√y4=yx.|x||y2|=yx.xy2=1yyx.x2y4=yx.x2y4=yx.|x||y2|=yx.xy2=1y.
(Do x>0x>0 nên
a,\(\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{y^4}}=\dfrac{y}{x}\cdot\dfrac{|x|}{|y^2|}=\dfrac{y}{x}\cdot\dfrac{x}{y^2}=\dfrac{1}{y}\)
\(2y^2\cdot\sqrt{\dfrac{x^4}{4y^2}}=2y^2\cdot\dfrac{\sqrt{x^4}}{\sqrt{4y^2}}=2y^2\cdot\dfrac{|x^2|}{|2y|}=2y^2\cdot\dfrac{x^2}{-2y}=\dfrac{2y^2\cdot x^2}{-2y}=-x^2y\)
\(5xy\sqrt{\dfrac{25x^2}{y^6}}=5xy\cdot\dfrac{\sqrt{25x^2}}{\sqrt{y^6}}=5xy\cdot\dfrac{|5x|}{|y^3|}=5xy\cdot\dfrac{-5x}{y^3}=\dfrac{5xy\cdot\left(-5x\right)}{y^3}=\dfrac{25x^2}{y^2}\)
\(0,2x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}=0,2x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}=\dfrac{2}{10}\cdot x^3y^3\cdot\dfrac{4}{|x^2y^4|}=\dfrac{1}{5}\cdot x^3\cdot y^3\cdot\dfrac{4}{x^2y^4}=\dfrac{4x^3y^3}{5x^2y^4}=\dfrac{4x}{5y}\)
a)1/y
b)-x2y
c)-25x2
d)4x/5y
a) = 1/y
b) = x^2y
c) = 2x^2/y^2
d) = 4x/5y
a,\(\dfrac{y}{x}\).\(\sqrt{\dfrac{x^2}{y^4}}\)=\(\dfrac{y}{x}\).\(\dfrac{x}{y^2}\)=\(\dfrac{1}{y}\)
b,\(2y^2\).\(\sqrt{\dfrac{x^4}{4y^2}}\)=\(2y^2\).\(\dfrac{x^2}{-2y}\)=-yx^2
c,5xy.\(\sqrt{\dfrac{25x^2}{y^6}}\)=5xy.\(\dfrac{-5x}{y^3}\)=\(\dfrac{-25x^2}{y^2}\)
d,\(0,2x^3y^3\).\(\sqrt{\dfrac{16}{x^4y^8}}\)=\(\dfrac{2}{10}x^3y^3\).\(\dfrac{4}{x^2y^4}\)=\(\dfrac{4x}{5y}\)
.
a) yx.√x2y4=yx.√x2√y4=yx.|x||y2|=yx.xy2=1y.
(Do x>0 nên |x|=x, y≠0 ⇒ y2>0 nên |y2|=y2)
b) 2y2.√x44y2=2y2.√x4√4y2=2y2.
Đúng(0)
a) \dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{\sqrt{x^2}}{\sqrt{y^4}}=\dfrac{y}{x}.\dfrac{|x|}{|y^2|}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{1}{y}xy.y4x2=xy.y4x2=xy.∣y2∣∣x∣=xy.y2x=y1.
(Do x>0x>0 nên |x|=x∣x∣=x, y \ne 0y
Đúng(0)
(Vì x > 0 nên |x| = x; y2 > 0 với mọi y ≠ 0)
(Vì x2 ≥ 0 với mọi x; và vì y < 0 nên |2y| = – 2y)
(Vì x < 0 nên |5x| = – 5x; y > 0 nên |y3| = y3)
(Vì x2y4 = (xy2)2 > 0 với mọi x ≠ 0, y ≠ 0)
Rút gọn các biểu thức sau:
a. \(\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}\) với x > 0; \(y\ne0;\) b. \(2y^2.\sqrt{\dfrac{x^4}{4y^2}}\) với y < 0;
c. \(5xy.\sqrt{\dfrac{25x^2}{y^6}}\) với x < 0, y > 0; d. \(0,2x^3y^3.\sqrt{\dfrac{16}{x^4y^8}}\) với \(x\ne0;y\ne0.\)
a)
= 
.
= 
.
= 
vì x > 0.
Do đó
= 
.
b)
= 
.
= 
.
.
Vì y < 0 nên │y│= -y. Do đó
= 
.
= 
.
c) 5xy.
= 5xy.
= 5xy.
.
Vì x < 0, y > 0 nên
= -x và 
= 
.
Do đó: 5xy
= 5xy.
= -
.
d) 0,2
= 
= 0,2
Rút gọn các biểu thức sau:
a) \(\frac{y}{x}\).\(\sqrt{\frac{x^2}{y^4}}\) vơi x>0;y\(\ne\)0
b) 2y\(^2\).\(\sqrt{\frac{x^4}{4y^2}}\) Y<0
c) 5xy.\(\sqrt{\frac{25x^2}{y^6}}\)với x<0,y>0
d) 0,2x\(^3\)y\(^3\).\(\sqrt{\frac{16}{x^4y^8}}\) với x\(\ne\)0,y\(\ne\)0
a/ \(\frac{y}{x}.\left(\sqrt{\frac{x^2}{y^4}}\right)=\frac{y}{x}.\frac{x}{y^2}=\frac{1}{y}\)
b/ \(2y^2.\sqrt{\frac{x^4}{4y^2}}=2y^2.\sqrt{\frac{\left(x^2\right)^2}{\left(-2y\right)^2}}=2y^2.\frac{x^2}{-2y}=-y.x^2\)
c/ \(5xy.\sqrt{\frac{25x^2}{y^6}}=5xy.\sqrt{\frac{\left(-5x\right)^2}{\left(y^3\right)^2}}=5xy.\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\)
d/\(0,2.x^3y^3.\sqrt{\frac{4^2}{\left(x^2y^4\right)^2}}=\frac{1}{5}.x^3y^3.\frac{4}{x^2y^4}=\frac{4x}{5y}\)
Trần Việt Linh sai phần b,c,d r bn
Sửa lại:
b) 2y\(^2\).\(\sqrt{\frac{x^4}{4y^2}}\) với y<0
Ta có : 2y\(^2\).\(\sqrt{\frac{x^4}{4y^2}}\)=2y\(^2\).\(\frac{x^2}{\left|y\right|}\)
Vì y>0 nên |y| = -y.Ta có : 2y\(^2\).\(\frac{x^2}{2\left|y\right|}\)= -2y\(^2\).\(\frac{x^2}{2y}\) = -2x\(^2\)y
c) 5xy.\(\sqrt{\frac{25x^2}{y^6}}\) với x<0,y>0
Ta có :5xy\(\sqrt{\frac{25x^2}{y^6}}\)=5xy.\(\frac{5\left|x\right|}{y^3}\) ( y>0)
Vì x<0 nên |x| =-x .Ta có : 5xy.\(\frac{5\left|x\right|}{y^3}\)= -5xy.\(\frac{5x}{y^3}\) =\(\frac{-25x^2}{y^2}\)
d) 0,,2x\(^3\)y\(^3\).\(\sqrt{\frac{16}{x^4y^8}}\) với x#o,y#0
Ta có: 0,2x\(^3\)y\(^3\)\(\frac{4}{x^2y^4}\)=\(\frac{0,8x}{y}\) ( vì #0,y#0)
Bài 47 (trang 27 SGK Toán 9 Tập 1)
Rút gọn:
a) $\dfrac{2}{x^{2}-y^{2}} \sqrt{\dfrac{3(x+y)^{2}}{2}}$ với $x \ge 0, y \ge 0$ và $x \ne y$ ;
b) $\dfrac{2}{2 a-1} \sqrt{5 a^{2}\left(1-4 a+4 a^{2}\right)}$ với $a>0,5$.
a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)
\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)
\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)
\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)
a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)
do \(x\ge0;y\ge0\)
\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)
bài 34:rúi gọn các biểu thức sau
a)\(\dfrac{2}{a}\sqrt{\dfrac{16a^2}{9}}\) (với a<0)b b)\(\dfrac{3}{a-1}\sqrt{\dfrac{4a^2-8a+4}{25}}\) (với a>1)
c)\(2\dfrac{y}{x}\sqrt{\dfrac{x^2}{4y^4}}\) (với y\(\ne\)0) d)\(\left(x-y\right)\sqrt{\dfrac{36}{x^2-2xy+y^2}}\) (với x\(\ne\)y)
éo biết
CM:
\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{6}}{6}\)
\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}+\sqrt{y}}=x-y\) với x.0, y>0, x≠y
\(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{x}}{y-\sqrt{xy}}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)với x>0, y>0, x≠y
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
Bài 20 (trang 15 SGK Toán 9 Tập 1)
Rút gọn các biểu thức sau:
a) $\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}$ với $a\ge 0$ ; b) $\sqrt{13a}.\sqrt{\dfrac{52}{a}}$ với $a>0$ ;
c) $\sqrt{5a}.\sqrt{45a}-3a$ với $a\ge 0$ ; d) $(3-a)^2-\sqrt{0,2}.\sqrt{180a^2}$.
a, \(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\left|\frac{a}{2}\right|=\frac{a}{2}\)
do \(a\ge0\)
b, \(\sqrt{13a}.\sqrt{\frac{52}{a}}=\sqrt{\frac{676a}{a}}=\sqrt{676}=26\)
c, \(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\left|15a\right|-3a\)
\(=15a-3a=12a\)do a > 0
d, \(=\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)
\(=\left(3-a\right)^2-\sqrt{36a^2}=\left(3-a\right)^2-\left|6a\right|\)
Với \(a\ge0\Rightarrow\left(3-a\right)^2-6a=a^2-6a+9-6a=a^2-12a+9\)
Với \(a< 0\Rightarrow\left(3-a\right)^2+6a=a^2-6a+9+6a=a^2+9\)
a) Ta có:
b) Ta có:
c) Do a ≥ 0 nên bài toán luôn xác định. Ta có:
d) Ta có:
Rút gọn
A=\(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
B=\(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}-\dfrac{x-y}{\sqrt{x}—\sqrt{y}}\)với\(x\ne y\)x>0;y>0
C=\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
D=\((3\sqrt{2}+\sqrt{6})\sqrt{6-3\sqrt{3}}\)
a: \(A=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}=10\)
b: \(B=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}=-2\sqrt{y}\)
c: \(C=\dfrac{\sqrt{3}-1}{\sqrt{6}-\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
Rút gọn
M=\(\dfrac{\sqrt{x}}{\sqrt{x}-6}\)-\(\dfrac{3}{\sqrt[]{x}+6}\)+\(\dfrac{x}{36-x}\)với x>0;x≠36
N=\(\dfrac{x^2\sqrt{xy}}{y}\).\(\sqrt{\dfrac{y}{x}}\)-x2 với x,y>0
P=\(\left(\dfrac{2-a\sqrt{a}}{2-\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{2-\sqrt{a}}{2-a}\right)\)2 với a>0;a≠2; ≠4
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
Bài 19 (trang 15 SGK Toán 9 Tập 1)
Rút gọn các biểu thức sau:
a) $\sqrt{0,36.a^2}$ với $a<0$ ; b) $\sqrt{a^4.(3-a)^2}$ với $a \ge 3$ ;
c) $\sqrt{27.48.(1-a)^2}$ với $a>1$ ; d) $\dfrac{1}{a-b}.\sqrt{a^4.(a-b)^2}$ với $a>b$.
Bạn học tốt nhé
a)0,6.a
b)\(a^2\).(a-3)
c)36.(a-1)
d)\(\dfrac{1.a^2}{a-b}\).(a-b)